Waveguide open-circuit?
Waveguide open-circuit?
(OP)
I am trying to work out how to calibrate a return loss measurement jig in waveguide. The system is just a CW power source, two directional couplers and two power meters. I thought that I could put on a short-circuit piece and an open circuit piece and say that both should give 0dB return loss. But then I realised that a waveguide "open-circuit" probably has no meaning. Just leaving the guide open presumably allows a large amount of field to escape. What would the return loss of an open waveguide be?
In order to get the traditional open and short circuit return loss values, does one simply use a sliding short or perhaps a short and a quarter lambda spacer?
In order to get the traditional open and short circuit return loss values, does one simply use a sliding short or perhaps a short and a quarter lambda spacer?





RE: Waveguide open-circuit?
There are variable waveguide shorts. They are made with a High Z/Low Z rod going down the centerline (forming a lowpass filter, with the lowpass corner frequency below the cuttoff of the waveguide) that can be moved by a micrometer up and down in the z axis.
But, since waveguide flanges are pretty repeatable, I would just craft up two or three waveguide fixed shorts of different length and use those. If it is just a scalar measurement, you could just stick on a different length short circuit when the machine asked for an "open". If it was a vector machine, you would have to program in the differening physical lengths to the short with some custom software.
The very old HP waveguide reflectometers were calibrated with a moving waveguide load to find the smith chart center, and a shorting plate to establish the phase reference, but I know of no modern network analyzers set up to do that sort of error correction today.
RE: Waveguide open-circuit?
Any idea what the return loss of the open waveguide would be?
RE: Waveguide open-circuit?
There are waveguide cal kits, check with Agilent and Maury Microwave.
kch
RE: Waveguide open-circuit?
Of course if I know the return loss of the open guide is 6dB (into RAM in the distance for example) that gives me another return loss check point.
6dB return loss, so 75% of the power is transmitted.
RE: Waveguide open-circuit?